Evaluate: `int_(-(3pi)/2)^(-pi/2)[(x+pi)^3+cos^2(x+3pi)]dx`

Answer

`pi//2`

Explanation

Let `I=int_(-3pi//2)^(-pi//2)[(x+pi)^(3)+cos^(2)(x+3pi)]dx`
Put `x=pi=t`, so that `dx=dt`
when `x=-(3pi)/2, t=-(pi)/2`
When `x=-(pi)/2,t=(pi)/2`
`:.I=int_(-pi//2)^(pi//2) [t^(3)+cos^(2)(t+2pi)]dt`
`=int_(-pi//2)^(pi//2) [t^(3)+cos^(2)t]dt`
`=0+2int_(0)^(pi//2) cos^(2) t dt =int_(0)^(pi//2)(1+cos2t)dt`
`=(pi)/2+0=(pi)/2`